<header>
    泰勒公式
</header>
<p>
    如果函数ƒ在点x<sub>0</sub>可导，则有
    <span class="oneline">
        ƒ(x)=ƒ(x<sub>0</sub>)+ƒ<sup>'</sup>(x<sub>0</sub>)(x-x<sub>0</sub>)+o(x-x<sub>0</sub>)
    </span>
    也就是在点x<sub>0</sub>附件，用一次多项式ƒ(x<sub>0</sub>)+ƒ<sup>'</sup>(x<sub>0</sub>)(x-x<sub>0</sub>)逼近函数ƒ(x)的时的误差是(x-x<sub>0</sub>)的高阶无穷小。
</p>
<p>
    现在考虑任意n次多项式
    <span class="oneline">
        p<sub>n</sub>(x)=a<sub>0</sub>+a<sub>1</sub>(x-x<sub>0</sub>)+a<sub>2</sub>(x-x<sub>0</sub>)<sup>2</sup>+ ...
        +a<sub>n</sub>(x-x<sub>0</sub>)<sup>n</sup>
    </span>
    逐次求它在点x<sub>0</sub>处的各阶导数，得到
    <span class="oneline">
        p<sub>n</sub>(x<sub>0</sub>)=a<sub>0</sub> ,
        p<sup>'</sup><sub>n</sub>(x<sub>0</sub>)=a<sub>1</sub> ,
        p<sup>"</sup><sub>n</sub>(x<sub>0</sub>)=2!a<sub>2</sub> , ... ,
        p<sup>(n)</sup><sub>n</sub>(x<sub>0</sub>)=n!a<sub>n</sub>
    </span>
    即
    <span class="oneline">
        a<sub>0</sub>=p<sub>n</sub>(x<sub>0</sub>) ,
        a<sub>1</sub>=<code>["division",["join",["rightBottom",["rightTop","p","'"],"n"],"(",["rightBottom","x","0"],")"],"1!"]</code>
        ,
        a<sub>2</sub>=<code>["division",["join",["rightBottom",["rightTop","p","\""],"n"],"(",["rightBottom","x","0"],")"],"2!"]</code>
        , ... ,
        a<sub>n</sub>=<code>["division",["join",["rightBottom",["rightTop","p","(n)"],"n"],"(",["rightBottom","x","0"],")"],"n!"]</code>
    </span>
    由此可见，多项式p<sub>n</sub>(x)的各项系数由其在点x<sub>0</sub>的各阶导数值所唯一确定。
</p>
<p>
    <span class="title">
        定理（带有佩亚诺型余项的泰勒公式）
    </span>
    若函数ƒ在点x<sub>0</sub>存在直到n阶导数，则有：
    <span class="oneline">
        ƒ(x)=ƒ(x<sub>0</sub>)+f<sup>'</sup>(x<sub>0</sub>)(x-x<sub>0</sub>)+
        <code>["division",["join",["rightTop","ƒ","\""],"(",["rightBottom","x","0"],")"],"2!"]</code>(x-x<sub>0</sub>)<sup>2</sup>
        + ... +
        <code>["division",["join",["rightTop","ƒ","(n)"],"(",["rightBottom","x","0"],")"],"n!"]</code>(x-x<sub>0</sub>)<sup>n</sup>
        +
        o((x-x<sub>0</sub>)<sup>n</sup>)
    </span>
</p>